cosecA+1/cosecA-1=(secA+tanA) ²
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Given that
LHS = [tanA/secA+1] + [cotaA/cosecA+1]
= tan A(sec.A — 1)/(sec A+1)(sec A—1) + cot.A(cosec.A — 1)/(cosec A+1)(cosecA — 1)
= tanA(secA — 1)/sec^2 A — 1 + cot.A(cosec.A — 1)/cosec^2A — 1
= tanA(secA — 1)/tan^2A + cot.A(cosec.A — 1)/cot^2A
= sec A-1/tanA+ cosec A — 1/ cotA
= sec A/tan A – 1/tan A + cosec A / cot A – 1 / cot A
= cosec A + sec A – [1/tan A + 1/cot A]
= cosec A + sec A – [cos^2A + sin^2A] / [sin A cos A]
= cosec A + sec A – sec A cosec A = RHS
Hence, proved.
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